How to find a complete expression for the derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$

The graph of the function is:


The derivative function is

$$\frac{dy}{dx}=\begin{cases}\frac{2}{1+x^2}&\text{if }|x|\le 1\\\frac{-2}{1+x^2}&\text{if }|x|>1\end{cases}$$

My Attempt

Fullsimplify[D[ArcSin[(2*x)/(1 + x^2)], x]]



PowerExpand[-((2 Sqrt[(-1 + x^2)^2/(1 + x^2)^2])/(-1 + x^2))]



Why am I not getting the other part $\frac{+2}{1+x^2}$ ?.

How do I evaluate the derivative in cases like this ?

  • 4
    $\begingroup$ Does FullSimplify[D[ArcSin[(2 x)/(1 + x^2)], x], x ∈ Reals] suffice? $\endgroup$ – Carl Woll Apr 6 '18 at 18:08
  • $\begingroup$ @CarlWoll for the given function we have both +ve and -ve derivatives, so i'd like to split the square root. Is it possible to do that?. and why does mathematica gives only one part of the solution for what i entered ? $\endgroup$ – ss1729 Apr 6 '18 at 18:18
  • 1
    $\begingroup$ @ss1729 PowerExpand ignores the fact that the square-root stands for positive square-root. Hence you lose that information in the simplified result. Carl Woll's solution is able to preserve that information using the signum function. $\endgroup$ – Subho Apr 6 '18 at 18:47
  • $\begingroup$ @Subho95 ok thanx. Carl Woll's suggestion is giving $$ \frac{\frac{-4 x^2}{(1 + x^2)^2} + \frac{2}{1+x^2}}{\sqrt{1 - \frac{4 x^2}{(1 + x^2)^2}}} $$, but from that how do I reach the form $\frac{\pm 2x}{1+x^2}$ ? $\endgroup$ – ss1729 Apr 6 '18 at 19:20
  • $\begingroup$ @ss1729 This is strange. You should directly get $\frac{2* Sign[-1 + x^2]}{(1 + x^2)}$ $\endgroup$ – Subho Apr 6 '18 at 19:26


PiecewiseExpand[FullSimplify[D[ArcSin[(2 x)/(1 + x^2)], x], x ∈ Reals], x ∈ Reals]

enter image description here

  • 1
    $\begingroup$ Assuming[Element[x, Reals], D[ArcSin[2 x/(1 + x^2)], x] // Simplify // PiecewiseExpand] $\endgroup$ – Bob Hanlon Apr 6 '18 at 20:13
  • $\begingroup$ Thanx. So whats with $x\in \text{Reals}$ ?. How is it different ? $\endgroup$ – ss1729 Apr 6 '18 at 21:11
  • 1
    $\begingroup$ @ss1729. FullSimplify[D[ArcSin[(2 x)/(1 + x^2)], x], x \[Element] Reals] and now try:FullSimplify[D[ArcSin[(2 x)/(1 + x^2)], x], x \[Element] Complexes] and what conclusions ? $\endgroup$ – Mariusz Iwaniuk Apr 6 '18 at 21:53

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